Question

part b - real - life applications

1. a tree is 30 feet tall. a support cable is attached from the top of the tree to a point 34 feet from the base of the tree. how long is the cable?
2. a ship sails 60 km north and then 25 km east. how far is the ship from its starting point?
3. a construction worker leans a 50 ft beam against a wall so that it touches the wall at 48 ft high. how far is the bottom of the beam from the wall?
4. a rectangular television has a diagonal of 55 inches and a width of 48 inches. what is the height of the television?

Explanation:

Step1: Identify right - triangle and Pythagorean theorem

For each problem, we can form a right - triangle and use the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(c\) is the hypotenuse and \(a\) and \(b\) are the other two sides.

Step2: Solve problem 6

Let the height of the tree be \(a = 30\) feet and the distance from the base of the tree be \(b=34\) feet. The length of the cable \(c\) is given by \(c=\sqrt{a^{2}+b^{2}}=\sqrt{30^{2}+34^{2}}=\sqrt{900 + 1156}=\sqrt{2056}=2\sqrt{514}\approx45.34\) feet.

Step3: Solve problem 7

The ship sails 60 km north (\(a = 60\) km) and 25 km east (\(b = 25\) km). The distance from the starting point \(c\) is \(c=\sqrt{a^{2}+b^{2}}=\sqrt{60^{2}+25^{2}}=\sqrt{3600+625}=\sqrt{4225}=65\) km.

Step4: Solve problem 8

The length of the beam is \(c = 50\) ft and the height on the wall is \(a = 48\) ft. Let the distance of the bottom of the beam from the wall be \(b\). Then \(b=\sqrt{c^{2}-a^{2}}=\sqrt{50^{2}-48^{2}}=\sqrt{(50 + 48)(50 - 48)}=\sqrt{98\times2}=\sqrt{196}=14\) ft.

Step5: Solve problem 9

The diagonal of the television is \(c = 55\) inches and the width is \(b = 48\) inches. Let the height be \(a\). Then \(a=\sqrt{c^{2}-b^{2}}=\sqrt{55^{2}-48^{2}}=\sqrt{(55 + 48)(55 - 48)}=\sqrt{103\times7}=\sqrt{721}\approx26.85\) inches.

Answer:

1. Approximately \(45.34\) feet
2. \(65\) km
3. \(14\) ft
4. Approximately \(26.85\) inches